Question

How many moles are in 937 g $Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}$?

Answer 1

To solve this question, we first need to understand what is a mole. Mole is the SI unit of measurement and is used to determine the amount of a substance. A mole of any substance has exactly $6.022\times {{10}^{23}}$ particles which can be ions, atoms, electrons, or molecules.

Complete step-by-step answer: We know that if one mole of a substance is present, it has exactly the Avogadro number $({{N}_{A}})$ of particles.
${{N}_{A}}=6.022\times {{10}^{23}}$
Now, the mass of a sample can be given by the sum of the mass of all the particles in it.
So, we can say that the mass of one mole of a compound is equivalent to the mass of all the particles contained in one mole of a substance i.e., $6.022\times {{10}^{23}}$ particles.
The mass of one mole of a substance is known as the molar mass of that substance. Its SI base unit is kg/mol but it is usually expressed in g/mol. It is a bulk property of a substance, not a molecular property.
Now, the molar mass of a sample is given by the mass of the sample substance divided by the number of moles of the substance present in the sample.
$M=\dfrac{m}{n}$
So, the number of moles in a given sample can be given by
$n=\dfrac{m}{M}$
Where n is the number of moles,
m is the mass of the substance given (in grams), and
M is the molar mass of the substance (in g/mol).
Now, the molecular mass of one mole of a compound is equal to the sum of atomic masses of all the elements present in the molecule.
So, the molecular mass of one mole of $Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}$ is

& {{M}_{Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}}}={{M}_{Ca}}+2[2\times {{M}_{C}}+3\times {{M}_{H}}+2\times {{M}_{O}}] \\\ & {{M}_{Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}}}=40.078+2[2\times 12.0107+3\times 1.00784+2\times 15.999] \\\ & {{M}_{Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}}}\cong 158.17g/mol \\\ \end{aligned}$$ It is given to us that the mass of the sample substance is 937 grams. So, the number of moles of the substance in the sample is $$\begin{aligned} & n=\dfrac{937g}{158.17g/mol} \\\ & n\cong 5.924mol \\\ \end{aligned}$$ **Hence there are approximately 5.924 moles in a sample of 937 grams of calcium acetate ($$Ca{{({{C}_{2}}{{H}_{3}}{{O}_{2}})}_{2}}$$).** **Note:** It must be noted that in 2019, the SI base unit of molar mass was redefined. According to the new definition, the molar mass constant is $${{M}_{u}}=0.99999999965\times {{10}^{-3}}kg/mol$$ And not $1\times {{10}^{-3}}kg/mol$. But since the change is so insignificant, for practical purposes, the molar mass of an element is still considered to be equivalent to the atomic mass of the element.